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In `a` solution the concentration of `CaCl_(2)` is `5M &` that of `MgCl_(2)` is `5 m`. The specific gravity of solution is `1.05`, calculate the concentration of `Cl^(-)` in the solution in terms of Molarity. |
Answer» Correct Answer - `[Cl^(-)]=13.36M` `CaCl_(2)rarr5M=555g` in `1L` solution or in `1050g` solution `wt`. Of `("solvant" + MgCl_(2))=1050-555=495g` `MgCl_(2)rarr5m` `1000g` solvant `rarr 5` mol of `MgCl_(2)` `=5xx95=475gMgCl_(2)` `i.e., 1475 ("solvant" + MgCl_(2))rarr 475g MgCl_(2)` `495("solvant"+ MgCl_(2)) rarr(475)/(1475)xx495` `=159.4g MgCl_(2)` moles of `MgCl_(2)=(159.4)/(95)=1.678` Total moles of `Cl^(-)` `=(5+1.678)xx2=13.356` volume of solution `1L` Molarity of `Cl^(-) = 13.356 M` |
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